Question
The pressure of a nitrogen gas, N2, cylinder is measured at 250 K and found to be 4.0 atm. What is the pressure of the same cylinder at room temperature, 293 K?
Answers
There are no human answers yet.
Answered by
GPT 3.5
To solve this problem, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Since we are looking for the final pressure at room temperature, we can write the equation as:
P1V1/T1 = P2V2/T2
Given:
P1 = 4.0 atm
V1 = unknown (since it does not change)
T1 = 250 K
P2 = unknown
V2 = unknown (since it does not change)
T2 = 293 K
We can rearrange the equation to solve for P2:
P2 = P1 * (T2 / T1)
Substitute the given values into the equation:
P2 = 4.0 atm * (293 K / 250 K)
P2 = 4.0 atm * 1.172
P2 = 4.688 atm
Therefore, the pressure of the same nitrogen gas cylinder at room temperature (293 K) would be 4.688 atm.
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Since we are looking for the final pressure at room temperature, we can write the equation as:
P1V1/T1 = P2V2/T2
Given:
P1 = 4.0 atm
V1 = unknown (since it does not change)
T1 = 250 K
P2 = unknown
V2 = unknown (since it does not change)
T2 = 293 K
We can rearrange the equation to solve for P2:
P2 = P1 * (T2 / T1)
Substitute the given values into the equation:
P2 = 4.0 atm * (293 K / 250 K)
P2 = 4.0 atm * 1.172
P2 = 4.688 atm
Therefore, the pressure of the same nitrogen gas cylinder at room temperature (293 K) would be 4.688 atm.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.